We have a recursively defined sequence a_{n}. a_{0}=0,a_{1}=3, and a_{n

Ava-May Nelson 2021-08-16 Answered
We have a recursively defined sequence \(\displaystyle{a}_{{{n}}}\).
\(\displaystyle{a}_{{{0}}}={0},{a}_{{{1}}}={3}\), and \(\displaystyle{a}_{{{n}}}={3}{a}_{{{n}-{1}}}-{2}{a}_{{{n}-{2}}}\) for \(\displaystyle{n}\geq{2}\)
We would like to prove that f or all \(\displaystyle{n}\geq{0},{a}_{{{n}}}={3}\cdot{2}^{{{n}}}-{3}\).
Prove this using the stronger mathematical induction.

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Expert Answer

escumantsu
Answered 2021-08-17 Author has 9249 answers
Step 1
Given that: \(\displaystyle{\left\lbrace{a}_{{{n}}}\right\rbrace}\) be a sequance of defined recursively \(\displaystyle{a}_{{{0}}}={0},\ {a}_{{{1}}}={3}\)
\(\displaystyle{a}_{{{n}}}={3}{a}_{{{n}-{1}}}-{2}{a}_{{{n}-{2}}}\) for \(\displaystyle{n}\geq{2}\)
To show: \(\displaystyle{a}_{{{n}}}={3}\cdot{2}^{{{n}}}-{3}\) for \(\displaystyle{n}\geq{2}\)
we will use strong mathematical induction
for \(\displaystyle{n}={0}\ {a}_{{{0}}}={3}\cdot{2}^{{{0}}}-{3}={0}\)
\(\displaystyle{n}={1}\ {a}_{{{1}}}={3}\cdot{2}^{{{1}}}-{3}={3}\)
So, for \(\displaystyle{n}={0},\ {n}={1}\) it is true
Let us assume it is true for all \(\displaystyle{k}\leq{n}\)
Such that \(\displaystyle{a}_{{{k}}}={3}\cdot{2}^{{{k}}}-{3}\ \forall{k}\leq{n}\)
Now for \(\displaystyle{k}={n}+{1}\)
So, \(\displaystyle{a}_{{{n}+{1}}}={3}{a}_{{{n}}}-{2}{a}_{{{n}-{1}}}\)
\(\displaystyle={3}{\left({3}\cdot{2}^{{{n}}}-{3}\right)}-{2}{\left({3}\cdot{2}^{{{n}-{1}}}-{3}\right)}\)
\(\displaystyle{\left(\because\right.}\) by induction \(\displaystyle{a}_{{{n}}}={3}\cdot{2}^{{{n}}}-{3},\ {a}_{{{n}-{1}}}={3}\cdot{2}^{{{n}-{1}}}-{3}{)}\)
\(\displaystyle{a}_{{{n}+{1}}}={3}^{{{2}}}\cdot{2}^{{{n}}}-{9}-{6}\cdot{2}^{{{n}-{1}}}+{6}\)
\(\displaystyle={3}\cdot{2}^{{{n}-{1}}}{\left({3}\cdot{2}-{2}\right)}-{3}\)
\(\displaystyle={3}\cdot{2}^{{{n}-{1}}}\cdot{4}-{3}\)
\(\displaystyle={3}\cdot{2}^{{{n}-{1}}}\cdot{2}^{{{2}}}-{3}\)
\(\displaystyle{a}_{{{n}+{1}}}={3}\cdot{2}^{{{n}+{1}}}-{3}\)
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